Answer:
(a) convex mirror
(b) virtual and magnified
(c) 23.3 cm
Explanation:
The having mirror is convex mirror.
distance of object, u = - 20 cm
magnification, m = 1.4
(a) As the image is magnified and virtual , so the mirror is convex in nature.
(b) The image is virtual and magnified.
(c) Let the distance of image is v.
Use the formula of magnification.

Use the mirror equation, let the focal length is f.

Radius of curvature, R = 2 f = 2 x 11.67 = 23.3 cm
Answer:
c.
Explanation:
We are given that
Acceleration due to gravity on the moon=
Acceleration due to gravity on the earth=

Net force due to am on an object on moon=
There is no friction and no drag force and there is no gravity involved
Then, the force acting on an object on earth=
(given)


Hence, option c is true.
Answer:
3.135 kN/C
Explanation:
The electric field on the axis of a charged ring with radius R and distance z from the axis is E = qz/{4πε₀[√(z² + R²)]³}
Given that R = 58 cm = 0.58 m, z = 116 cm = 1.16m, q = total charge on ring = λl where λ = charge density on ring = 180 nC/m = 180 × 10⁻⁹ C/m and l = length of ring = 2πR. So q = λl = λ2πR = 180 × 10⁻⁹ C/m × 2π(0.58 m) = 208.8π × 10⁻⁹ C and ε₀ = permittivity of free space = 8.854 × 10⁻¹² F/m
So, E = qz/{4πε₀[√(z² + R²)]³}
E = 208.8π × 10⁻⁹ C × 1.16 m/{4π8.854 × 10⁻¹² F/m[√((1.16 m)² + (0.58 m)²)]³}
E = 242.208 × 10⁻⁹ Cm/{35.416 × 10⁻¹² F/m[√(1.3456 m² + 0.3364 m²)]³}
E = 242.208 × 10⁻⁹ Cm/35.416 × 10⁻¹² F/m[√(1.682 m²)]³}
E = 6.839 × 10³ Cm²/[1.297 m]³F
E = 6.839 × 10³ Cm²/2.182 m³F
E = 3.135 × 10³ V/m
E = 3.135 × 10³ N/C
E = 3.135 kN/C
Answer:
7.74m/s
Explanation:
Mass = 35.9g = 0.0359kg
A = 39.5cm = 0.395m
K = 18.4N/m
At equilibrium position, there's total conservation of energy.
Total energy = kinetic energy + potential energy
Total Energy = K.E + P.E
½KA² = ½mv² + ½kx²
½KA² = ½(mv² + kx²)
KA² = mv² + kx²
Collect like terms
KA² - Kx² = mv²
K(A² - x²) = mv²
V² = k/m (A² - x²)
V = √(K/m (A² - x²) )
note x = ½A
V = √(k/m (A² - (½A)²)
V = √(k/m (A² - A²/4))
Resolve the fraction between A.
V = √(¾. K/m. A² )
V = √(¾ * (18.4/0.0359)*(0.395)²)
V = √(0.75 * 512.53 * 0.156)
V = √(59.966)
V = 7.74m/s
B is the answer that I know of.